frictionnewtonian-mechanicsrotational-dynamicstorque
In a horizontal surface, a block (cube) is sliding due to a sudden push. When the block slides, there is frictional force which is acting on the block.
Frictional force will have a torque around the center of mass, so why does the block not rotate/roll around (a horizontal axis through) the center of mass instead of sliding?
Your understanding that a change in rotational velocity requires a net torque (calculated around any convenient axis) is correct.
The only forces acting are gravity, the normal force from the ramp, and the force of friction between the ramp and the block. Choosing an axis across the ramp through the center of mass of the block makes life simpler; the force of gravity (all of it) acts through this point, and thus exerts no torque.
Next, consider the friction force, that is given by $mg \sin \theta$. This acts up the ramp, at the point of contact of the ramp and block, and thus exerts a torque around the center of mass depending on the dimensions of the block. (Tall blocks are more tippy than squat blocks) In this case, the torque is a clockwise one.
In order for there to be no rotation, there must be a counter-clockwise torque about the CofM. This means that the diagram is wrong; the normal force no longer is distributed uniformly along the base, and cannot be ignored as acting through the CofM. In fact, the normal force shifts towards the front of the block as the ramp slope increases, and when it reaches the front corner, the rear corner lifts and the block tumbles...
A lot of material in your question. Let me try to summarize:
You have an cube sliding down a frictionless slope of 45°, when it suddenly encounters a surface with friction $\mu$. Will it topple?
The answer is "yes" if the sum of two forces - gravity and deceleration - points along a line that does not pass through the base of the cube. Since the cube was on a 45 degree slope in your example, the line of action of the force of gravity is already pointing at the corner of the cube:
If you add any friction, that will result in an additional torque on the block that will topple it.
The problem is more interesting when you have a more stable situation - lower angle of slope (or wider block): in that case, the question of whether the block will topple (assuming that the coefficient of friction is unchanged with contact area - which is a reasonable assumption) has a more complex answer (but it's not what you asked for).
Best Answer
First consider the initial push. If the friction is high enough and the push is high enough on the block it will roll instead of slide. Generally static friction is greater than dynamic friction, so if it starts sliding it will continue.
If the friction gets great enough, the block can roll. Say you have a box sliding on ice that comes to the edge of the ice and starts sliding on dirt. The friction will increase considerably. You compare the torque the friction applies to the box against the torque required to lift the back edge against gravity. If it is greater, the box will start to roll.